I am trying to search based on multiple conditions which works but the problem is that does not behave like this.
Assuming i have a search query like
Orders.find({$or: {"status":{"$in":["open", "closed"]},"paymentStatus":{"$in":["unpaid"]}}}
)
and i add another filter parameter like approvalStatus it does not leave the previously found items but rather it treats the query like an AND that will return an empty collection of items if one of the queries does not match.
How can i write a query that regardless of what is passed into it, it will retain previously found items even if there is no record in one of the conditions.
like a simple OR query in sql
I hope i explained this well enough
Using $or here is the right approach, but its value needs to be an array of query expressions, not an object.
So your query should look something like this instead:
Orders.find({$or: [
{"status": {"$in": ["open", "closed"]}},
{"paymentStatus": {"$in": ["unpaid"]}},
{"approvalStatus": {"$in": ["approved"]}}
]})
Related
In my collection "users", each user is defined by their name, list of hobbies, and skills:
Structure of a document in "users":
name : "Name"
hobbies: ["hobby1","hobby2","hobby3"]
skills: ["skill1","skill2","skill3"]
What I want is: Get all documents in the collection where one of those fields contains a specific word. The problem is that currently I get all documents and then I do a loop to retain the matching documents. But I would prefer to do a query instead (reads are expensive). So I was thinking to reorganize the document like this:
New structure:
data: ["Name","hobby1","hobby2","hobby3",skill1","skill2","skill3"]
So I could do a query with "where" and "array-contains". What do you think? Isn't it strange?
If you want to have a OR statement between the hobbies and skills then you can't use a query that uses array-contains. As stated on this documentation:
You can use at most one array-contains clause per query. You can't combine array-contains with array-contains-any.
What I would suggest is to combine the skills and hobbies into one.:
e.g.:
name : "Name"
skillsandhobbies : ["Hobby1", "Hobby2", "Hobby3", "Skill1", "Skill2", "Skill3"]
Then you can use this sample query to get all the documents that have hobbies or skills that contain a specific word:
const q = query(collection(db, "users"), where("skillsandhobbies", "array-contains", "specific-word"));
As I understood from your comments, you basically need to perform a successive search. You need to search the hobbies array for a particular word, and if you don't get any results, you need to perform another search, in the skills array, for the exact same word, and if again you don't get any results, you need to search for the name.
Since you didn't specify a particular programming language, I assume you're using Flutter, as in the other questions of yours. So I will try to explain real quickly how to do it.
First, you need to create the following query:
final queryHobbiesByWord = users.where("hobbies", arrayContains: "word");
Then you have to call get(). As soon as you get a successful callback, you have to check if the number of documents that exist inside the QuerySnapshot is greater than zero. If it is, then perform the logic you want, otherwise, inside the callback create another query that looks like this:
final querySkillsByWord = users.where("skills", arrayContains: "word");
Do the exact same thing as about. If the number of documents that exist inside the QuerySnapshot is greater than zero, then perform the logic you want, otherwise, inside the callback create another query that looks like this:
final queryNameByWord = users.where("name", isEqualTo: "word");
Again, perform the same operation as above, and that's pretty much it. Don't forget to perform each operation inside the callback of the previous one.
my question is very similar to how-to-get-multiple-document-using-array-of-mongodb-id, however, I would like to find multiple documents without using the _id.
That is, consider that I have documents such as
the
document = { _id: _id, key_1: val_1, key_2: val_2, key_3: val_3}
I need to be able to .find() by multiple parameters, as for example,
query_1 = {key_1: foo, key_2: bar}
query_2 = {key_1: foofoo, key_2: barbar}
Right now, I am running a query for query_1, followed by a query for query_2.
As it turns out, this method is extremely inefficient.
I tried to add concurrency as to make it faster, but the speedup was not even 2x.
Is it possible to query multiple documents at once?,
I am looking for a method that returns the union of the matches for query_1 AND query_2.
If this is not possible, do you have any suggestions that might speed a query of this type?
Thank you for your help.
I have a mongodb collection full of 65k+ documents, each one with a properties named site_histories. The value of it is an array that might be empty, or might not be. If it is not empty, it will have one or more objects similar to this:
"site_histories" : "[{\"site_id\":\"129373\",\"accepted\":\"1\",\"rejected\":\"0\",\"pending\":\"0\",\"user_id\":\"12743\"}]"
I need to make a query that will look for every instance in the collection of a document that does not have a given user_id.
I'm pretty new to Mongo, so I was trying to make a query that would find every instance that does have the given user_id, which I was then planning on adding a "$ne" to, but even that didn't work. This is the query I was using that didn't work:
db.test.find({site_histories: { $elemMatch: {user_id: '12743\' }}})
So can anyone tell me why this query didn't work? And can anyone help me format a query that will do what I need the final query to do?
If your site_histories really is an array, it should be as simple as doing:
db.test.find({"site_histories.user_id": "12743"})
That looks in all the elements of the array.
However, I'm a bit scared of all those backslashes. If site_histories is a string, that won't work. It would mean that the schema is poorly designed, you'd maybe try with $regex
What I'm trying to do:
Filter a field of a collection that matches a given condition. Instead of returning every item in the field (which is an array of items), I only want to see matched items.
Similar to
select items from test where items.histPrices=[10,12]
It is also similar to what's found on the mongodb website here: http://www.mongodb.org/display/DOCS/Retrieving+a+Subset+of+Fields
Here's what I have been trying:
db.test.save({"name":"record", "items":[{"histPrices":[10,12],"name":"stuff"}]})
db.test.save({"name":"record", "items":[{"histPrices":[10,12],"name":"stuff"},
{"histPrices":[12,13],"name":"stuff"},{"histPrices":[11,14],"name":"stuff"}]})
db.test.find({},{"name":1,"items.histPrices":[10, 12]})
It will return all the objects that have a match for items.histPrices:[10,12], including ALL of the items in items[]. But I don't want the ones that don't match the condition.
From the comments left on Mongodb two years ago, the solution to get only the items with that histPrices[10,12] is to do it with javascript code, namely, loop through the result set and filter out the other items.
I wonder if there's a way to do that with just the query.
Your find query is wrong
db.test.find({},{"name":1,"items.histPrices":[10, 12]})
Your condition statement should be in the first part of the find statement.In your query {} means fetch all documents similar to this sql
select items from test (no where clause)
you have to change your mongodb find to
db.test.find({"items.histPrices":[10, 12]},{"name":1})
make it work
since your items is an array and if you wanted to return only the matching sub item, you have to use positional operator
db.test.find({"items.histPrices":[10, 12]},{"name":1,'items.$':1})
When working with arrays Embedded to the Document, the best approach is the one suggested by Chien-Wei Huang.
I would just add another aggregation, with the $group (in cases the document is very long, you may not want to retrieve all its content, only the array elements) Operator.
Now the command would look like:
db.test.aggregate({$match:{name:"record"}},
{$unwind:"$items"},
{$match {"items.histPrices":[10, 12]}},
{$group: {_id: "$_id",items: {$push: "$items"}}});)
If you are interested to return only one element from the array in each collection, then you should use projection instead
The same kind of issue solved here:
MongoDB Retrieve a subset of an array in a collection by specifying two fields which should match
db.test.aggregate({$unwind:"$items"}, {$match:{"items.histPrices":[10, 12]}})
But I don't know whether the performance would be OK. You have to verify it with your data.
The usage of $unwind
If you want add some filter condition like name="record", just add another $march at first, ex:
db.test.aggregate({$match:{name:"record"}}, {$unwind:"$items"}, {$match:{"items.histPrices":[10, 12]}})
https://jira.mongodb.org/browse/SERVER-828
Get particular element from mongoDB array
MongoDB query to retrieve one array value by a value in the array
I am making an application using pymongo wrapper for which my schema is like:
{
_id: <some_id>,
name: <some_name>,
my_tags: [<list_of_tags>]
}
Now I want to return those entries which falls under the user specified tags. For example,
I want to have entries where my_tags should be atleast ["college", "USA", "engineering"]. For that I read $all construct can be used. Now what I want to know is, would it be of any use making an index on my_tags. For my app, this type of queries are used extensively.
would it be of any use making an index on my_tags. For my app, this type of queries are used extensively.
Yes $all will use an index so it is still good to make one there however there are still optimisations that can be done for it: https://jira.mongodb.org/browse/SERVER-5331 and https://jira.mongodb.org/browse/SERVER-1000
Normally the docs will only warn you of when something can not use an index.
The syntax for the $all query is:
db.collection.find({'my_tags': {'$all': ['college', 'USA', 'engineering']}})
The documentation can be found at:
http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24all